Calculus of a Single Variable, Chapter 2, 2.1, Section 2.1, Problem 35

The given line is :-
3x - y + 1 = 0
or, y = 3x + 1 (the line is represented in slope intercept form)
Thus, the slope of the line = 3
Now, the tangent to the curve f(x) = (x^3) is parallel to the above line
Thus, the slope of the tangent = slope of the line = 3.......(1)
The given function is:-
f(x) = (x^2)
differentiating both sides w.r.t 'x' we get
f'(x) = 3(x^2)
Now, slope of the tangent = 3
Thus, 3(x^2) = 3
or, x = +1 or -1
Putting the value of x =1 in the given equation of curve, we get
f(1) = y = 1
Hence the tangent passes through the point (1,1)
Thus, equation of the tangent at the point (1,1) and having slope = 3 is :-
y - 1 = (3)*(x - 1)
or, y - 1 = 3x - 3
or, y - 3x + 2 = 0 is the equation of the tangent to the given curve at (1,1).........(2)

Putting x = -1 in the equation of curve we get
f(-1) = y = -1
Thus, the equation of the tangent passing through the point (-1,-1) and having slope 3 will be:-
y - (-1) = 3(x - (-1))
or, y + 1 = 3x + 3
or, y - 3x - 2 = 0